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Failure probability prior to attaining MTTF

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by

Murray Wiseman
in

The article “What is the scale parameter” showed that 63% of randomly failing items or items adhering to Weibull failure behavier will fail prior to attaining their MTTF. Generally speaking, the MTTF is not a good decision guide for the age replacement of an item. The following shows that for age dependent failure behavior (e.g. shape factor  β =2) over 50% of the items will fail prior to the MTTF.

The article “MTTF is the area under the reliability curve” showed that:

\int_{0}^{\infty}R(t)dt=\int_{0}^{\infty}t(f(t))dt=E(t)=MTTF

Assuming  Weibull behavior:

(1)   \begin{align*} MTTF & =\int R\left ( t \right )dt \nonumber \\ & = \int e^{-\left ( \frac{t}{\eta } \right )^{\beta }}dt \\ & =\int e^{-\left ( \frac{1}{\eta } \right )^{\beta }\left ( t \right )^{\beta }}dt \end{align*}

Substitute β=2 and a=\left ( \frac{1}{\eta } \right )^{\beta }

(2)   \begin{align*} MTTF & =\int e^{-at^{2}}dt \nonumber \\ \end{align*}

Knowing that \int e^{-ax^{2 }}dx=\frac{1}{2}\left ( \frac{\pi }{a} \right )^{\frac{1}{2}}, then

(3)   \begin{align*} MTTF & = \frac{1}{2}\left ( \frac{\pi }{a} \right )^{\frac{1}{2}}\nonumber \\ & = \frac{1}{2}\left ( \frac{\pi }{\left ( \frac{1}{\eta } \right )^{2}} \right )^{\frac{1}{2}} \\ & = 0.89\eta \end{align*}

At t=MTTF = .89η

(4)   \begin{align*} F(t) & = 1-e^{-\left ( \frac{t}{\eta } \right )^{\beta }}\nonumber \\ & = 1-\frac{1}{e^{.89^{2}}} \\ & = 0.547107279 \end{align*}

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Comments

4 responses to “Failure probability prior to attaining MTTF”

  1. Decisiones de Mantenimiento |

    […] common measure of reliability is the Mean Time To Failure (MTTF). It can be calculated from the probability distribution. We show in the TBM discussion that, given the ages of failure […]

  2. What is the scale parameter? |

    […] Not only those that fail randomly but all that adhere to Weibull behavior (estimated at 80+%). See Failure probability prior to attaining MTTF […]

  3. Daming Lin Avatar
    Daming Lin

    Don’t need to be that complicated. Just use the well-known formula for the Weibull mean: eta*gamma(1+1/beta).

    So you apply the formula F(MTTF)=1-exp(-gamma(1+1/beta)^beta) to get the Cumulative Failure Probability at t=MTTF for any beta.

    1. Murray Wiseman Avatar
      Murray Wiseman

      Thanks, Daming.

      So using Excel formula =GAMMA(1+1/2), sure enough we get 0.886227 yielding the same cumulative failure probability F(MTTF)=55%

      For beta = 3 we get MTTF=0.89298 yielding a cumulative failure probability of =1-1/EXP(0.89298^3) = 51%

      For the above the scale parameter (eta) = 1 can be used since horizontal axis need not be scaled to fit any particular data sample.

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